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VLD [36.1K]
2 years ago
11

Rearrange the kinetic energy equation so that velocity is the subject

Physics
1 answer:
mars1129 [50]2 years ago
6 0

Answer: V = sqrt(2K.E/M)

Explanation:

K.E = 1 /2MV^2

MV^2 = 2 K.E

V^2 = 2K.E /M

V = sqrt(2K.E/M)

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The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the
8090 [49]
Compute first for the vertical motion, the formula is:

y = gt²/2 

0.810 m = (9.81 m/s²)(t)²/2 

t = 0.4064 s 


whereas the horizontal motion is computed by: 

x = (vx)t 

4.65 m = (vx)(0.4064 s) 

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed. 

(vy) = gt 

(vy) = (9.81 m/s²)(0.4064 s) 

(vy) = 3.99 m/s 


speed with which it hit the ground: 

v = sqrt[(vx)² + (vy)²] 

v = sqrt[(11.44 m/s)² + (3.99 m/s)²] 

v = 12.12 m / s
6 0
3 years ago
A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball
lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.  

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

M_{1} u_{1} +M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

Since initial velocity for M1 ball is zero, then

M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

and

M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

So, on solving all the above equation, we get an equation for velocity and that is

\frac{2M_{2}u_{2} }{(M_{1}+M_{2}  }=final velocity of ball with mass 2.5 kg

v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is \frac{1}{2}*2.5*(4.67)^{2}  =27.09 J

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

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300kg of water are lifted 10m vertically in 5s show the work done in 30kj and that power is 6kw . Please help me​
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Answer:

6KW

Explanation:

The computation is shown below:

We know that

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Here

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And

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The above represent the answer

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